Law of Cosines: Videos & Practice Problems

To solve for a side in a triangle when given certain angles and sides, the Law of Sines is often a go-to method. However, there are situations where this approach is not applicable, particularly when you have two sides but lack the angles opposite those sides, or when you have an angle but not its corresponding side. In such cases, the Law of Cosines becomes essential.

The Law of Cosines provides a way to relate the lengths of the sides of a triangle to the cosine of one of its angles. The general formula for the Law of Cosines is expressed as:

$$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$$

In this equation, \(c\) is the side opposite angle \(C\), while \(a\) and \(b\) are the other two sides. This formula allows you to calculate the length of side \(c\) when you know the lengths of sides \(a\) and \(b\) and the measure of angle \(C\).

For example, if you have \(a = 3\), \(b = 2\), and \(C = 60^\circ\), you can substitute these values into the Law of Cosines:

$$c^2 = 3^2 + 2^2 - 2 \cdot 3 \cdot 2 \cdot \cos(60^\circ)$$

Calculating this gives:

$$c^2 = 9 + 4 - 12 \cdot \frac{1}{2}$$

$$c^2 = 13 - 6 = 7$$

Taking the square root results in:

$$c = \sqrt{7}$$

This method not only allows you to find the lengths of sides but can also be used to determine missing angles if all three sides are known. It's important to note that the Law of Cosines encompasses the Pythagorean theorem as a special case, where the angle is \(90^\circ\). Understanding these relationships and formulas is crucial for solving various triangle problems effectively.

In solving triangles, the law of cosines is essential for determining unknown sides and angles, particularly in SAS (Side-Angle-Side) and SSS (Side-Side-Side) configurations. When given two sides and the included angle in an SAS triangle, the law of cosines allows us to find the third side. The formula used is:

$$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$$

In this context, if we have a triangle where side a is 4, side b is 3, and angle C is 60 degrees, we can substitute these values into the formula to find side c:

$$c^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \cos(60^\circ)$$

Calculating this gives:

$$c^2 = 16 + 9 - 2 \cdot 4 \cdot 3 \cdot \frac{1}{2}$$

$$c^2 = 25 - 12 = 13$$

Thus, c is equal to the square root of 13, or approximately 3.6. However, it is often preferable to leave answers in radical form to avoid rounding errors.

Once the third side is found, the next step is to determine one of the remaining angles using the law of cosines again. For angle A, the formula is:

$$A^2 = b^2 + c^2 - 2bc \cdot \cos(A)$$

Substituting the known values:

$$4^2 = 3^2 + 13 - 2 \cdot 3 \cdot \sqrt{13} \cdot \cos(A)$$

Rearranging and simplifying leads to:

$$16 = 9 + 13 - 6\sqrt{13} \cdot \cos(A)$$

From here, we can isolate cos(A):

$$-6\sqrt{13} \cdot \cos(A) = -6$$

Thus, we find:

$$\cos(A) = \frac{1}{\sqrt{13}}$$

Taking the inverse cosine gives us angle A as approximately 73.9 degrees. Since the cosine function yields only one angle in the range of 0 to 180 degrees, no further calculations for a second angle are necessary.

Finally, to find the last angle B, we use the angle sum property of triangles:

$$A + B + C = 180^\circ$$

Substituting the known values:

$$73.9 + B + 60 = 180$$

Solving for B gives:

$$B = 180 - 73.9 - 60 = 46.1^\circ$$

In conclusion, the triangle's dimensions are complete with side c as the square root of 13, angle A as 73.9 degrees, and angle B as 46.1 degrees. Mastery of the law of cosines is crucial for effectively solving these types of triangle problems.